如何在python 2.7和python 3x中以时间戳（13位-毫秒）格式考虑上个月的第一天和上个月的最后一天？

Question

Suppose today is August 28, 2019 i want timestamp of July 1,12:00:00 AM and July 31,23:59:59 PM . In general, I want the previous month date and time irrespective of year.

Answer 1

The following should work for both Python 3x and Python2.7

from datetime import date
from dateutil.relativedelta import relativedelta

today = date.today()
d = today - relativedelta(months=1)

first_day = date(d.year, d.month, 1)
first_day.strftime('%A %d %B %Y')
#returns first date of the previous month - datetime.date(2019, 7, 1)

last_day = date(today.year, today.month, 1) - relativedelta(days=1)
last_day.strftime('%A %d %B %Y')

# returns the last date of the previous month - datetime.date(2019, 7, 31)

Note - The above code only returns the first and last date, not the time. I don't really understand why can't you just hard code the time in milliseconds since it will always be fixed, irrespective of the date.

Answer 2

Alternatively with no additional modules:

from datetime import datetime, timedelta
now = datetime.now()
print(now)
firstOfThisMonth = now.replace(day=1)
print(firstOfThisMonth)
endOfLastMonth = firstOfThisMonth-timedelta(days=1)
print(endOfLastMonth)
firstOfLastMonth = endOfLastMonth.replace(day=1)
print(firstOfLastMonth)
print(firstOfLastMonth.replace(hour=0, minute=0, second=0, microsecond=0))
print(endOfLastMonth.replace(hour=23, minute=59, second=59, microsecond=0))

Outputs the following

2019-08-28 07:36:38.768223
2019-08-01 07:36:38.768223
2019-07-31 07:36:38.768223
2019-07-01 07:36:38.768223
2019-07-01 00:00:00
2019-07-31 23:59:59